Question

g The fencing of the left border costs $10 per foot, while the fencing of the lower border costs $2 per foot. (No fencing is required along the river.) You want to spend $520 and enclose as much area as possible. What are the dimensions of your garden, and what area does it enclose?

Answer 1

Answer:

$Area = 1690ft^2\\x = 130ft, y = 26ft$

Step-by-step explanation:

Given the costs we can form an equation:

$10y + 2x =520 - Eq(A)$

and fencing is triangular such that the area enclosed can be written as:

$A = \dfrac{xy}{2} -Eq(B)$

• First need to convert the above equation so that it is only in terms of one variable. [either x or y]

To make the equation only in terms of $x$ we can substitute $y$ from Eq(A) i.e, $y =\frac{520-2x}{10}$, to Eq(B)

$A = \dfrac{x}{2} \dfrac{520-2x}{10}$

simplify

$A = \dfrac{520x - 2x^2}{20}$

$A = -\dfrac{1}{10}x^2 + 26x$

• Now, in order to find the maximum area enclosed we can find $\frac{dA}{dx}$ and equate it zero.

$\dfrac{dA}{dx} = -\dfrac{1}{5}x + 26$

$0 = -\dfrac{1}{5}x + 26$

$-26 = -\dfrac{1}{5}x$

$x = 130$

we have the length of one dimension: specifically, the lower fence will be $x =130ft$

we can use this value of $x$ to find the corresponding value of $y$. From Eq(A)

$10y + 2x =520$

$10y +2(130) = 520$

$y = \dfrac{520 - 2(130)}{10}$

$y = 26$

the length of the left fence will be $y =26ft$

• The enclosed area by the fence will be

$A = \dfrac{xy}{2}$

$A = \dfrac{(130)(26)}{2}$

$A = 1690$

Hence the maximum area that can enclosed by the fences provided the costs will be $1690ft^2$

• You can even check the cost of the dimensions whether they all add up to $520 or not.

Use Eq(A)

$10y + 2x =520$

$10(26) + 2(130) =520$

and indeed it does!